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13 de jun. de 2020 · It's dangerous to declare a non-return statement to a non-void function (that's why the warning) , your function as int-return must have an integer output. So, add return total; or change your function into a void , knowing that total is an attribute, so it can be considered as a mutator.
29 de fev. de 2012 · While the shortviewed solution is to return a value, it is better if a pop() is of type void. template <typename T> void Stack<T>::pop() { ... } The reason is that you cannot write an exception safe pop()-function that returns something: template <typename T> T Stack<T>::pop() { T foo = stor_.top(); // May throw.
- The function signature: template T Stack ::pop() tells the compiler that your function will return a type T however your function...
- You must add the return keyword: template T Stack ::pop() { return myStack.pop_front(); } template T Stack ::peek() con...
- This will work: template void Stack ::pop() { `myStack.pop_front(); } template void Stack ::pee...
- Your signature asks you to return an object of type T. template T Stack ::pop() To break this down a bit template means th...
- As you signature indicate that function is returning type T, you should change your code to code like following for both function. template
11 de nov. de 2022 · Change the function to not return anything (void). Check all your functions that are supposed to return a value and make sure that they do return a value.
Omitting the return statement in a function which is has a return type that is not void is undefined behavior. int function() { // Missing return statement } int main() { function(); //Undefined Behavior }
30 de abr. de 2010 · If you have a C++ program missing a return statement from a function that is supposed to return a value, g++ will compile it happily with no errors (or even a warning, unless -Wreturn-type or -Wall is used). Trying to use the return value from the function will most likely cause a segmentation fault.
4 de out. de 2018 · You can always use -Werror=return-type, or -fsanitize=undefined which will diagnose it at runtime.
23 de abr. de 2024 · Setting the return type of the function to void makes the sketch behave as expected. My conclusion is that the situation is not as clear cut as it could be
